What is the integral of cos(aln(x/b)) ?

The integral of cos(aln(x/b)) is a/(a^2+1)xsin(aln(x/b))+1/(a^2+1)xcos(aln(x/b))+C

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integral of cos(aln(x/b))

\int cos (a ln (\frac{x}{b})) d x

\frac{a}{a ^{2} + 1} x sin (a ln (\frac{x}{b})) + \frac{1}{a ^{2} + 1} x cos (a ln (\frac{x}{b})) + C

Solution steps

\int cos (a ln (\frac{x}{b})) d x

Apply u-substitution

: \int \frac{e ^{\frac{u}{a}} b cos ( u )}{a} d u

= \int \frac{e ^{\frac{u}{a}} b cos ( u )}{a} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{b}{a} \cdot \int e^{\frac{u}{a}} cos (u) d u

Apply u-substitution

: \int e^{v} cos (v a) a d v

= \frac{b}{a} \cdot \int e^{v} cos (v a) a d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{b}{a} a \cdot \int e^{v} cos (v a) d v

Apply Integration By Parts

: e^{v} \frac{1}{a} sin (v a) - \int e^{v} \frac{1}{a} sin (v a) d v

= \frac{b}{a} a (e^{v} \frac{1}{a} sin (v a) - \int e^{v} \frac{1}{a} sin (v a) d v)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{b}{a} a (e^{v} \frac{1}{a} sin (v a) - \frac{1}{a} \cdot \int e^{v} sin (v a) d v)

Apply Integration By Parts

: - e^{v} \frac{1}{a} cos (a v) - \int - e^{v} \frac{1}{a} cos (a v) d v

= \frac{b}{a} a (e^{v} \frac{1}{a} sin (v a) - \frac{1}{a} (- e^{v} \frac{1}{a} cos (a v) - \int - e^{v} \frac{1}{a} cos (a v) d v))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{b}{a} a (e^{v} \frac{1}{a} sin (v a) - \frac{1}{a} (- e^{v} \frac{1}{a} cos (a v) - (- \frac{1}{a} \cdot \int e^{v} cos (a v) d v)))

Therefore

\frac{b}{a} \cdot \int e^{v} cos (a v) d v a = \frac{b}{a} a (e^{v} \frac{1}{a} sin (v a) - \frac{1}{a} (- e^{v} \frac{1}{a} cos (a v) - (- \frac{1}{a} \cdot \int e^{v} cos (a v) d v)))

Isolate

\int e^{v} cos (a v) d v

= \frac{b}{a} (\frac{a e ^{v} sin ( a v )}{a ^{2} + 1} + \frac{e ^{v} cos ( a v )}{a ^{2} + 1}) a

Substitute back

= \frac{b}{a} \frac{a e ^{\frac{a l n ( \frac{x}{b} )}{a}} sin ( a \frac{a l n ( \frac{x}{b} )}{a} )}{a ^{2} + 1} + \frac{e ^{\frac{a l n ( \frac{x}{b} )}{a}} cos ( a \frac{a l n ( \frac{x}{b} )}{a} )}{a ^{2} + 1} a

Simplify

\frac{b}{a} \frac{a e ^{\frac{a l n ( \frac{x}{b} )}{a}} sin ( a \frac{a l n ( \frac{x}{b} )}{a} )}{a ^{2} + 1} + \frac{e ^{\frac{a l n ( \frac{x}{b} )}{a}} cos ( a \frac{a l n ( \frac{x}{b} )}{a} )}{a ^{2} + 1} a : \frac{a}{a ^{2} + 1} x sin (a ln (\frac{x}{b})) + \frac{1}{a ^{2} + 1} x cos (a ln (\frac{x}{b}))

= \frac{a}{a ^{2} + 1} x sin (a ln (\frac{x}{b})) + \frac{1}{a ^{2} + 1} x cos (a ln (\frac{x}{b}))

Add a constant to the solution

= \frac{a}{a ^{2} + 1} x sin (a ln (\frac{x}{b})) + \frac{1}{a ^{2} + 1} x cos (a ln (\frac{x}{b})) + C

f (x) = \frac{1}{x - 2}

y^{^{'}} - 2 x (y^{2} + 1) = 0

\frac{d}{d x} ((4 x + 1)^{2} (1 - x)^{3})

x \to 0 lim (\frac{x + 1}{x + 4})

What is the integral of cos(aln(x/b)) ?
The integral of cos(aln(x/b)) is a/(a^2+1)xsin(aln(x/b))+1/(a^2+1)xcos(aln(x/b))+C